Question

Given $P(A \cup B ) = 0.6,P(A\cap B) = 0.2$ , the probability of exactly one of the event occurs is

• A. $0.4$
• B. $0.2$
• C. $0.6$
• D. $0.8$

Answer 1

Answer: (A)

$0.4$

Answer 2

Given, $P(A \cup B)=0.6, P(A \cap B)=0.2$ Probability of exactly one of the event occurs is $P(\bar{A} \cap B)+P(A \cap \bar{B})$ $=P(B)-P(A \cap B)+P(A)-P(A \cap B)$ $=P(A \cup B)+P(A \cap B)-2 P(A \cap B)$ ${[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)]}$ $=P(A \cup B)-P(A \cap B)$ $=0.6-0.2$ $=0.4$